Physics, asked by Brainlyconquerer, 10 months ago

In the circuit shown in the figure there are two parallel plate capacitors each of capacitance C. The switch S1 is pressed first to fully charge the capacitor C1 and then released. The switch S2 is then pressed to charge the capacitor C2. After sometime S2 is released and then S3 is pressed. after sometime:

A) the charge on the upper plate of C1 is 2CV_0

B) the charge on the upper plate of C1 is CV_0

C) the charge on the upper plate of C2 is zero

D) the charge on the upper plate of C2 is minus -CV_0

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Answers

Answered by PopulistAnswerer
14

Physics:——

  • [S1]charge of upper plate of [C1] is [+2CV0]

  • Then pressing [S2] this charge equally distributes in two capacitors

  • hence charge an upper plates of both capacitors will be [+CV0]

  • Then after deleting the [S2] and [S3] is pressed.

  • Upper plate of [C1] unchanged (retained)

  • =[+CV0] but the thing is that when charge on upper plate of [C2] is from to New battery [=−CV0]

Also refer the attachment

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Answered by Aɾꜱɦ
7

Answer:

[S1]charge of upper plate of [C1] is [+2CV0]

[S1]charge of upper plate of [C1] is [+2CV0]Then pressing [S2] this charge equally distributes in two capacitors

[S1]charge of upper plate of [C1] is [+2CV0]Then pressing [S2] this charge equally distributes in two capacitorshence charge an upper plates of both capacitors will be [+CV0]

[S1]charge of upper plate of [C1] is [+2CV0]Then pressing [S2] this charge equally distributes in two capacitorshence charge an upper plates of both capacitors will be [+CV0]Then after deleting the [S2] and [S3] is pressed.

[S1]charge of upper plate of [C1] is [+2CV0]Then pressing [S2] this charge equally distributes in two capacitorshence charge an upper plates of both capacitors will be [+CV0]Then after deleting the [S2] and [S3] is pressed.Upper plate of [C1] unchanged (retained)

[S1]charge of upper plate of [C1] is [+2CV0]Then pressing [S2] this charge equally distributes in two capacitorshence charge an upper plates of both capacitors will be [+CV0]Then after deleting the [S2] and [S3] is pressed.Upper plate of [C1] unchanged (retained)=[+CV0] but the thing is that when charge on upper plate of [C2] is from to New battery [=−CV0]

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